A body of mass m1 = 1.5 kg moving along a directed axis in the positive sense with a velocity

Physics

1 answer:

larisa86 [58]3 years ago

8 0

Answer:

3.71 m/s in the negative direction

Explanation:

From collisions in momentum, we can establish the formula required here which is;

m1•u1 + m2•v2 = m1•v1 + m2•v2

Now, we are given;

m1 = 1.5 kg

m2 = 14 kg

u1 = 11 m/s

v1 = -1 m/s (negative due to the negative direction it is approaching)

u2 = -5 m/s (negative due to the negative direction it is moving)

Thus;

(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)

This gives;

16.5 - 70 = -1.5 + 14v2

Rearranging, we have;

16.5 + 1.5 - 70 = 14v2

-52 = 14v2

v2 = - 52/14

v2 = 3.71 m/s in the negative direction

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Firdavs [7]

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4 0

3 years ago

The induced magnetic field at radial distance 4.0 mm from the central axis of a circular parallel-plate capacitor is 1.8 ✕ 10−7

wel

Answer:

$\frac{dE}{dt}=2.07*10^{13}\frac{V/m}{s}$

Explanation:

According to Gauss's law, the electric flux through the circular plates is defined as the electric field multiplied by its area:

$\Phi=EA=E(\pi R^2)(1)$

The magnetic field around the varying electric field of the circular plates is given by:

$B=\frac{\epsilon_0 \mu_o}{2\pi r}\frac{d\Phi}{dt}(2)$

Replacing (1) in (2) and solving for $\frac{dE}{dt}$ :

$B=\frac{\epsilon_0 \mu_o\pi R^2}{2\pi r}\frac{dE}{dt}\\\frac{dE}{dt}=\frac{2rB}{\epsilon_0 \mu_o R^2}\\\frac{dE}{dt}=\frac{2(4*10^{-3}m)(1.8*10^{-7}T)}{(8.85*10^{-12}\frac{C^2}{N\cdot m^2})(4\pi *10{-7}\frac{Tm}{A})(2.5*10^{-3}m)^2}\\\\\frac{dE}{dt}=2.07*10^{13}\frac{V/m}{s}$